# invariant subspaces for self-adjoint *-algebras of operators

Let $H$ be a Hilbert space and $B(H)$ its algebra of bounded operators. Recall that, given an operator $T\in B(H)$, a subspace   $V\subseteq H$ is said to be invariant  for $T$ if $Tx\in V$ whenever $x\in V$.

Similarly, given a subalgebra $\mathcal{A}\subseteq B(H)$, we will say that a subspace $V\subseteq H$ is invariant for $\mathcal{A}$ if $Tx\in V$ whenever $T\in\mathcal{A}$ and $x\in V$, i.e. if $V$ is invariant for all operators in $\mathcal{A}$.

## Invariant subspaces for a single operator

Let $T\in B(H)$. If a subspace $V\subset H$ is invariant for $T$, then so is its closure   $\overline{V}$.

Proof: Let $x\in\overline{V}$. There is a sequence $\{x_{n}\}$ in $V$ such that $x_{n}\to x$. Hence, $Tx_{n}\to Tx$. Since $V$ is invariant for $T$, all $Tx_{n}$ belong to $V$. Thus, their limit $Tx$ must be in $\overline{V}$. We conclude that $\overline{V}$ is also invariant for $T$. $\square$

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Proposition 2 - Let $T\in B(H)$. If a subspace $V\subset H$ is invariant for $T$, then its orthogonal complement   $V^{\perp}$ is invariant for $T^{*}$.

Proof: Let $y\in V^{\perp}$. For all $x\in H$ we have that $\langle x,T^{*}y\rangle=\langle Tx,y\rangle=0$, where the last equality comes from the fact that $Tx\in V$, since $V$ is invariant for $T$. Therefore $T^{*}y$ must belong to $V^{\perp}$, from which we conclude that $V^{\perp}$ is invariant for $T^{*}$. $\square$

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Proposition 3 - Let $T\in B(H)$, $V\subset H$ a closed subspace and $P\in B(H)$ the orthogonal projection onto $V$. The following are statements are equivalent      :

1. 1.

$V$ is invariant for $T$.

2. 2.

$V^{\perp}$ is invariant for $T^{*}$.

3. 3.

$TP=PTP$.

Proof: $(1)\Longrightarrow(2)$ This part follows directly from Proposition 2.

$(2)\Longrightarrow(1)$ From Proposition 2 it follows that $(V^{\perp})^{\perp}$ is invariant for $(T^{*})^{*}=T$. Since $V$ is closed, $V=\overline{V}=(V^{\perp})^{\perp}$. We conclude that $V$ is invariant for $T$.

$(1)\Longrightarrow(3)$ Let $x\in H$. From the orthogonal decomposition theorem we know that $H=V\oplus V^{\perp}$, hence $x=y+z$, where $y\in V$ and $z\in V^{\perp}$. We now see that $TPx=Ty$ and $PTPx=PTy=Ty$, where the last equality comes from the fact that $Ty\in V$. Hence, $TP=PTP$.

$(3)\Longrightarrow(1)$ Let $x\in V$. We have that $Tx=TPx=PTPx$. Since $PTPx$ is obviously on the image of $P$, it follows that $Tx\in V$, i.e. $V$ is invariant for $T$. $\square$

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Proposition 4 - Let $T\in B(H)$, $V\subset H$ a closed subspace and $P\in B(H)$ the orhtogonal projection onto $V$. The subspaces $V$ and $V^{\perp}$ are both invariant for $T$ if and only if $TP=PT$.

Proof: $(\Longrightarrow)$ From Proposition 3 it follows that $V$ is invariant for both $T$ and $T^{*}$. Then, again from Proposition 3, we see that $PT=(T^{*}P)^{*}=(PT^{*}P)^{*}=PTP=TP$.

$(\Longleftarrow)$ Suppose $TP=PT$. Then $PTP=TPP=TP$, and from Proposition 3 we see that $V$ is invariant for $T$.

We also have that $PT^{*}=T^{*}P$, and we can conclude in the same way that $V$ is invariant for $T^{*}$. From Proposition 3 it follows that $V^{\perp}$ is also invariant for $T$. $\square$

## Invariant subspaces for *-algebras of operators

Proposition 5 - Let $\mathcal{A}$ be a *-subalgebra of $B(H)$ and $V$ a subspace of $H$. If a subspace $V$ is invariant for $\mathcal{A}$, then so are its closure $\overline{V}$ and its orthogonal complement $V^{\perp}$.

Proof: From Proposition 1 it follows that $\overline{V}$ is invariant for all operators in $\mathcal{A}$, which means that $V$ is invariant for $\mathcal{A}$.

Also, from Proposition 2 it follows that $V^{\perp}$ is invariant for the adjoint  of each operator in $\mathcal{A}$. Since $\mathcal{A}$ is self-adjoint, it follows that $V^{\perp}$ is invariant for $\mathcal{A}$. $\square$

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Let $\mathcal{A}$ be a *-subalgebra of $B(H)$, $V\subset H$ a closed subspace and $P$ the orthogonal projection onto $V$. The following are equivalent:

1. 1.

$V$ is invariant for $\mathcal{A}$.

2. 2.

$V^{\perp}$ is invariant for $\mathcal{A}$.

3. 3.

$P\in\mathcal{A}^{\prime}$, i.e. $P$ belongs to the commutant of $\mathcal{A}$.

Proof: $(1)\Longleftrightarrow(2)$ This equivalence follows directly from Proposition 5 and the fact that $V$ is closed.

$(1)\Longrightarrow(3)$ Suppose $V$ is invariant for $\mathcal{A}$. We have already proved that $V^{\perp}$ is also invariant for $\mathcal{A}$. Thus, from Proposition 4 it follows that $P$ commutes with all operators in $\mathcal{A}$, i.e. $P\in\mathcal{A}^{\prime}$.

$(3)\Longrightarrow(1)$ Suppose $P\in\mathcal{A}^{\prime}$. Then $P$ commutes with all operators in $\mathcal{A}$. From Proposition 4 it follows that $V$ is invariant for each operator in $\mathcal{A}$, i.e. $V$ is invariant for $\mathcal{A}$. $\square$

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